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Two copper, one steel rod: 3000 kg

Short answer

Equal shortening yields σ_steel = 1.60 σ_copper. Equilibrium then gives stress in each copper rod 66.31 kg/cm² and in the steel rod 106.10 kg/cm² (compressive).

Two copper rods and one steel rod are having diameter 4 cm, together support a load 3000 kg as shown in figure below. Determine the stresses in each rod. Take Es=2×106 kg/cm2E_s = 2\times10^6\ \text{kg/cm}^2, Ec=106 kg/cm2E_c = 10^6\ \text{kg/cm}^2. [Figure: a rigid plate carrying a 3000 kg load rests on two outer copper rods and a central steel rod standing on a base; the copper rods are 4 m long and the steel rod stands in a recess 1 m deeper.]

The load is carried by a rigid plate resting on the three rods, so all rods shorten equally. The steel rod stands in a recess 1 m deeper, so it is 5 m long (copper 4 m).

Compatibility

δc=δs⇒σcLcEc=σsLsEs⇒σc(400)106=σs(500)2×106⇒σs=1.60 σc\delta_c = \delta_s \Rightarrow \frac{\sigma_c L_c}{E_c} = \frac{\sigma_s L_s}{E_s} \Rightarrow \frac{\sigma_c(400)}{10^6} = \frac{\sigma_s(500)}{2\times10^6} \Rightarrow \sigma_s = 1.60\,\sigma_c

Equilibrium

Area of each rod A=π4(4)2=12.566A = \dfrac{\pi}{4}(4)^2 = 12.566 cm².

P=2σcA+σsA=σcA (2+1.60)P = 2\sigma_cA + \sigma_sA = \sigma_cA\,(2 + 1.60) 3000=σc(12.566)(3.60)⇒σc=66.315 kg/cm2,σs=106.103 kg/cm23000 = \sigma_c(12.566)(3.60) \Rightarrow \sigma_c = 66.315\ \text{kg/cm}^2,\quad \sigma_s = 106.103\ \text{kg/cm}^2

Check: 2(833.3)+1333.3=3000.02(833.3) + 1333.3 = 3000.0 kg.

Answer: stress in each copper rod = 66.31 kg/cm²; stress in the steel rod = 106.10 kg/cm² (compressive).

✓ Answer checked by Claude Opus 5.5 (high effort)

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